Use the Distance, Rate, and Time Formula

One formula you will use often in algebra and in everyday life is the formula for distance traveled by an object moving at a constant rate. Rate is an equivalent word for “speed.” The basic idea of rate may already familiar to you. Do you know what distance you travel if you drive at a steady rate of 60 miles per hour for 2 hours? (This might happen if you use your car’s cruise control while driving on the highway.) If you said 120 miles, you already know how to use this formula!

We will use the Strategy for Solving Applications that we used earlier in this chapter. When our problem requires a formula, we change Step 4. In place of writing a sentence, we write the appropriate formula. We write the revised steps here for reference.

You may want to create a mini-chart to summarize the information in the problem. See the chart in this first example.

Example 2.58

Jamal rides his bike at a uniform rate of 12 miles per hour for $3\frac{1}{2}$ hours. What distance has he traveled?

Solution

Step 1. Read the problem.
Step 2. Identify what you are looking for.		distance traveled
Step 3. Name. Choose a variable to represent it.		Let d = distance.
Step 4. Translate: Write the appropriate formula.		$d=rt$
Substitute in the given information.		$d=12·3\frac{1}{2}$
Step 5. Solve the equation.		$d=42$ miles
Step 6. Check
Does 42 miles make sense?
Jamal rides:
Step 7. Answer the question with a complete sentence.		Jamal rode 42 miles.

Example 2.59

Rey is planning to drive from his house in San Diego to visit his grandmother in Sacramento, a distance of 520 miles. If he can drive at a steady rate of 65 miles per hour, how many hours will the trip take?

Solution

Step 1. Read the problem.
Step 2. Identify what you are looking for.		How many hours (time)
Step 3. Name. Choose a variable to represent it.		Let t = time.
Step 4. Translate. Write the appropriate formula.		$d=rt$
Substitute in the given information.		$520=65t$
Step 5. Solve the equation.		$t=8$
Step 6. Check. Substitute the numbers into the formula and make sure the result is a true statement.
	$\begin{array}{ccc}\hfill d& =\hfill & rt\hfill \\ \hfill 520& \stackrel{?}{=}\hfill & 65·8\hfill \\ \hfill 520& =\hfill & 520✓\hfill \end{array}$
Step 7. Answer the question with a complete sentence. Rey’s trip will take 8 hours.

Solve a Formula for a Specific Variable

You are probably familiar with some geometry formulas. A formula is a mathematical description of the relationship between variables. Formulas are also used in the sciences, such as chemistry, physics, and biology. In medicine they are used for calculations for dispensing medicine or determining body mass index. Spreadsheet programs rely on formulas to make calculations. It is important to be familiar with formulas and be able to manipulate them easily.

In Example 2.58 and Example 2.59, we used the formula $d=rt$. This formula gives the value of $d$, distance, when you substitute in the values of $r\text{and}t$, the rate and time. But in Example 2.59, we had to find the value of $t$. We substituted in values of $d\text{and}r$ and then used algebra to solve for $t$. If you had to do this often, you might wonder why there is not a formula that gives the value of $t$ when you substitute in the values of $d\text{and}r$. We can make a formula like this by solving the formula $d=rt$ for $t$.

To solve a formula for a specific variable means to isolate that variable on one side of the equals sign with a coefficient of 1. All other variables and constants are on the other side of the equals sign. To see how to solve a formula for a specific variable, we will start with the distance, rate and time formula.

Example 2.61

Solve the formula $A=\frac{1}{2}bh$ for $h$:

ⓐ when $A=90$ and $b=15$ ⓑ in general

Solution

ⓐ when $A=90$ and $b=15$				ⓑ in general
Write the formula.				Write the formula.
Substitute.
Clear the fractions.				Clear the fractions.
Simplify.				Simplify.
Solve for $h$.				Solve for $h$.

We can now find the height of a triangle, if we know the area and the base, by using the formula $h=\frac{2A}{b}$.

The formula $I=Prt$ is used to calculate simple interest, I, for a principal, P, invested at rate, r, for t years.

Example 2.62

Solve the formula $I=Prt$ to find the principal, $P$:

ⓐ when $I=\text{\$}\mathrm{5,600},r=4\%,t=7\text{years}$ ⓑ in general

Solution

ⓐ $I=\mathrm{\$5,600}$, $r=\mathrm{4\%}$, $t=\mathrm{7\; years}$				ⓑ in general
Write the formula.				Write the formula.
Substitute.
Simplify.				Simplify.
Divide, to isolate P.				Divide, to isolate P.
Simplify.				Simplify.
The principal is

Later in this class, and in future algebra classes, you’ll encounter equations that relate two variables, usually x and y. You might be given an equation that is solved for y and need to solve it for x, or vice versa. In the following example, we’re given an equation with both x and y on the same side and we’ll solve it for y.

Example 2.63

Solve the formula $3x+2y=18$ for y:

ⓐ when $x=4$ ⓑ in general

Solution

ⓐ when $x=4$				ⓑ in general
Substitute.
Subtract to isolate the $y$-term.				Subtract to isolate the $y$-term.
Divide.				Divide.
Simplify.				Simplify.

In Examples 1.60 through 1.64 we used the numbers in part ⓐ as a guide to solving in general in part ⓑ. Now we will solve a formula in general without using numbers as a guide.

Example 2.64

Solve the formula $P=a+b+c$ for $a$.

Solution

We will isolate $a$ on one side of the equation.
Both $b$ and $c$ are added to $a$, so we subtract them from both sides of the equation.
Simplify.

Example 2.65

Solve the formula $6x+5y=13$ for y.

Solution

Subtract $6x$ from both sides to isolate the term with $y$.
Simplify.
Divide by 5 to make the coefficient 1.
Simplify.

The fraction is simplified. We cannot divide $13-6x$ by 5.